What is the integral of (x^4+x-4)/(x^2+2) ?

The integral of (x^4+x-4)/(x^2+2) is (x^3)/3-2x+1/2 ln|x^2+2|+C

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integral of (x^4+x-4)/(x^2+2)

\int \frac{x ^{4} + x - 4}{x ^{2} + 2} d x

\frac{x ^{3}}{3} - 2 x + \frac{1}{2} ln x^{2} + 2 + C

Solution steps

\int \frac{x ^{4} + x - 4}{x ^{2} + 2} d x

Long division

\frac{x ^{4} + x - 4}{x ^{2} + 2} : x^{2} - 2 + \frac{x}{x ^{2} + 2}

= \int x^{2} - 2 + \frac{x}{x ^{2} + 2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x - \int 2 d x + \int \frac{x}{x ^{2} + 2} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 2 d x = 2 x

\int \frac{x}{x ^{2} + 2} d x = \frac{1}{2} ln x^{2} + 2

= \frac{x ^{3}}{3} - 2 x + \frac{1}{2} ln x^{2} + 2

Add a constant to the solution

= \frac{x ^{3}}{3} - 2 x + \frac{1}{2} ln x^{2} + 2 + C

\frac{d}{d x} (\frac{x}{x ^{2} + 64})

\frac{d y}{d x} (x^{2}) = y^{3}

\int \frac{1}{3 t ^{2}} d t

\int \frac{e ^{a} + 1}{e ^{a}} d a

x \to - 8 lim (\frac{x ^{2} - 4}{8 - x})

What is the integral of (x^4+x-4)/(x^2+2) ?
The integral of (x^4+x-4)/(x^2+2) is (x^3)/3-2x+1/2 ln|x^2+2|+C